Moment of inertia of a thin rod of mass M and length L about an axis passing through centre is ML2/12. Its moment of inertia about a parallel axis at a distance of L/4 from the axis is given by?
A. ML2/48
B. ML3/48
C. ML2/12
D. 7ML2/48
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Solution
MI at the centre axis = ML² / 12 By using Parallel Axis Theorem = New MI = ML²/12 + M(L/4)² = 28ML² / 192 = 7ML² / 48